A 4-m square footing is shown in figure B which supports a load Q = 5000 kN. In this problem H1 = 8 m and H2 = 12 m. Calculate: a. The pressure at the base of the footing b. The pressure at the mid height of the clay layer assuming that the stress beneath the footing is spread at a slope of 1H to 2V c. The pressure at the mid height of the clay layer [below the center of the footing using Bousinessq equation assuming point load

Q: For the square footing shown below, B = 10 ft, qmax = 5 ksf and qmin = 1 ksf. a. Determine the value…

A: Solution:Given:The width of the square footing is B=10ftThe maximum intensity of contact pressure is…

Q: Two meters below the ground surface, a 2x2 m square footing is constructed on NC clayey sand. The…

A: Given data: Size of square footing = 2m * 2m Average Cone resistance = 20 MPa Layer of Hard rock is…

Q: A footing 2.25 x 2.25m is located at a depth of 1.5 m in a sand, the shear streng parameters to be…

A: Given : Footing size = 2.25 X 2.25 m Depth = 1.5 m To Find : Ultimate bearing capacity : A) if the…

Q: A square footing of 2 m side is subjected to a vertical load of 2000 kN. Assuming load dispersion as…

A:

Q: A column is supported on a footing as shown in the figure below. The water table is at a depth of 10…

A: Given- Water table at a depth of 10 m, Size of footing- 1.5 m × 3 m, Depth of footing Df=1 m, Unit…

Q: QI/ for the (3m*2m) footing shown in fig. below find the column load (v) that can be carried by the…

A: Given: Size of footing = 3×2 m ϕ=20°e=0.8Gs=2.7 c=30 kN/m2 γ=19 kN/m3 Factor of safety = 3 Depth of…

Methods to Determine Vertical Stress

The load applied to the soil is transferred to the underground. The presence of water at soil pores and solid grains above the soil are the primary components that are responsible for the effective load transfer. Due to the load acting, the internal component of the soil or the underground formations undergoes deformations or a tectonic shift. The amount of deformation depends upon the amount of load acting on the soil surface, which indeed, depends on the weight of everything present above the soil. Due to this, the underground soil element develops internal stresses which try to resist the tectonic changes.